Problem: Divide the following complex numbers. $ \dfrac{-14+5i}{1-4i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1+4i}$ $ \dfrac{-14+5i}{1-4i} = \dfrac{-14+5i}{1-4i} \cdot \dfrac{{1+4i}}{{1+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-14+5i) \cdot (1+4i)} {(1-4i) \cdot (1+4i)} = \dfrac{(-14+5i) \cdot (1+4i)} {1^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-14+5i) \cdot (1+4i)} {(1)^2 - (-4i)^2} = $ $ \dfrac{(-14+5i) \cdot (1+4i)} {1 + 16} = $ $ \dfrac{(-14+5i) \cdot (1+4i)} {17} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-14+5i}) \cdot ({1+4i})} {17} = $ $ \dfrac{{-14} \cdot {1} + {5} \cdot {1 i} + {-14} \cdot {4 i} + {5} \cdot {4 i^2}} {17} $ Evaluate each product of two numbers. $ \dfrac{-14 + 5i - 56i + 20 i^2} {17} $ Finally, simplify the fraction. $ \dfrac{-14 + 5i - 56i - 20} {17} = \dfrac{-34 - 51i} {17} = -2-3i $